How do you solve #4-p/(p-1)=2/(p-1)#?

Question

3202 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-17T07:18:02

#p=2#

#4-p/(p-1)=2/(p-1)# is equivalent to

#4=p/(p-1)+2/(p-1)#

or #4=(p+2)/(p-1)#

or #4xxcolor(red)((p-1))=(p+2)/(p-1)xxcolor(red)((p-1))#

or #4p-4=p+2#

or #4pcolor(red)(-p)-4color(red)(+4)=pcolor(red)(-p)+2color(red)(+4)#

or #3p=6#

or #3pxx1/3=6xx1/3=6/3=2#

i.e. #p=2#